miljanwpd2

2023-03-23

What is the projection of $<6,5,3>$ onto $<2,-1,8>$?

Saottah87e

Given $\stackrel{\to }{a}=<6,5,3>$ and $\stackrel{\to }{b}=<2,-1,8>$, we can find $pro{j}_{\stackrel{\to }{b}}\stackrel{\to }{a}$, the vector projection of $\stackrel{\to }{a}$ onto $\stackrel{\to }{b}$ using the following formula:
$pro{j}_{\stackrel{\to }{b}}\stackrel{\to }{a}=\left(\frac{\stackrel{\to }{a}\cdot \stackrel{\to }{b}}{|\stackrel{\to }{b}|}\right)\frac{\stackrel{\to }{b}}{|\stackrel{\to }{b}|}$
That is, the dot product of the two vectors divided by the magnitude of $\stackrel{\to }{b}$, multiplied by $\stackrel{\to }{b}$ divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide $\stackrel{\to }{b}$ by its magnitude in order to obtain a unit vector (vector with magnitude of $1$). You may have noticed that the first quantity is a scalar because we know that the dot product of two vectors always yields a scalar value.
Therefore, the scalar projection of $a$ onto $b$ is $com{p}_{\stackrel{\to }{b}}\stackrel{\to }{a}=\frac{a\cdot b}{|b|}$, also written $|pro{j}_{\stackrel{\to }{b}}\stackrel{\to }{a}|$
Taking the dot product of the two vectors will be our first step.
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=<6,5,3>\cdot <2,-1,8>$
$⇒\left(6\cdot 2\right)+\left(5\cdot -1\right)+\left(3\cdot 8\right)$
$⇒12-5+24=31$
Then, by taking the square root of the sum of the squares of each component, we can determine the size of $\stackrel{\to }{b}$
$|\stackrel{\to }{b}|=\sqrt{{\left({b}_{x}\right)}^{2}+{\left({b}_{y}\right)}^{2}+{\left({b}_{z}\right)}^{2}}$
$|\stackrel{\to }{b}|=\sqrt{{\left(2\right)}^{2}+{\left(-1\right)}^{2}+{\left(8\right)}^{2}}$
$⇒\sqrt{4+1+64}=\sqrt{69}$
And now we have everything we need to find the vector projection of $\stackrel{\to }{a}$ onto $\stackrel{\to }{b}$
$pro{j}_{\stackrel{\to }{b}}\stackrel{\to }{a}=\frac{31}{\sqrt{69}}\cdot \frac{<2,-1,8>}{\sqrt{69}}$
$⇒\frac{31<2,-1,8>}{69}$
$=\frac{31}{69}<2,-1,8>$
The coefficient can be distributed to each element of the vector and written as follows:
$⇒<\frac{62}{69},-\frac{31}{69},\frac{248}{69}>$
The scalar projection of $\stackrel{\to }{a}$ onto $\stackrel{\to }{b}$ is just the first half of the formula, where $com{p}_{\stackrel{\to }{b}}\stackrel{\to }{a}=\frac{a\cdot b}{|b|}$. Therefore, the scalar projection is $\frac{31}{\sqrt{69}}$, which does not simplify any further, besides to rationalize the denominator if desired, giving $\frac{31\sqrt{69}}{69}$

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